Answer:
[tex]f^{-1}(x)=\dfrac{1}{3}(\ln x +1)[/tex]
Step-by-step explanation:
Given function:
[tex]f(x)=e^{3x-1}[/tex]
Replace f(x) with y:
[tex]\implies y=e^{3x-1}[/tex]
Take natural logs of both sides:
[tex]\implies \ln y = \ln e^{3x-1}[/tex]
Apply the Power Log Law [tex]\ln x^n=n\ln x[/tex] :
[tex]\implies \ln y = (3x-1)\ln e[/tex]
As [tex]\ln e=1[/tex] then:
[tex]\implies \ln y = 3x-1[/tex]
Rearrange to make x the subject:
[tex]\implies 3x=\ln y +1[/tex]
[tex]\implies x=\dfrac{1}{3}(\ln y +1)[/tex]
Swap x for [tex]f^{-1}(x)[/tex] and y for x:
[tex]\implies f^{-1}(x)=\dfrac{1}{3}(\ln x +1)[/tex]
